On Fourier Series

Let $f:[-\pi, \pi] \rightarrow \mathbb{R}$ be defined as $f(x)=|x|$. Expand the function in a Fourier series and then evaluate the series:

$$a) \sum_{n={\rm odd} \geq 1}^{\infty}\frac{1}{n^2}, \; \; \;\;\;\;\;\;\;\; \;\;\;\;\;\;\;\; \beta)\sum_{n=1}^{\infty}\frac{1}{n^2}$$

Solution:

 
It is pretty straightforward to note that $f$ can be re-written as:
$$f(x)=\left\{\begin{matrix}
x &, x \in [0, \pi] \\
-x &, x \in [-\pi, 0)
\end{matrix}\right.$$

The function $f$ meets all Dirichlet's conditions , therefore it can be expanded in a Fourier Series of the form:

$$ f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left [ a_n \cos nx+ b_n \sin nx \right ]$$

All we have to do is to evaluate the coefficients. So , here goes:

$\color{grey} \blacksquare$     $\displaystyle a_0= \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\,{\rm d}x = \frac{1}{\pi}\int_{-\pi}^{\pi}\left | x \right |\,{\rm d}x=\pi $

$\color{grey} \blacksquare$     $\displaystyle a_n =\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\cos n x \, {\rm d}x =\frac{1}{\pi}\int_{-\pi}^{\pi}\left | x \right | \cos nx \,{\rm d}x =\cdots =\frac{2((-1)^n -1)}{\pi n^2}$

$\color{grey} \blacksquare$    $ \displaystyle {b_n = \frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\sin nx \,{\rm d}x = \require{cancel} \cancelto{0}{\frac{1}{\pi}\int_{-\pi}^{\pi}\left | x \right | \sin nx \,{\rm d}x} =0 }$

Hence:
$$f(x)=\frac{\pi}{2}+2 \sum_{n=1}^{\infty}\frac{(-1)^n -1}{\pi n^2}\cos n x$$

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$\mathbf{a)}$ We actually want to compute the series: $\displaystyle \sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$

The Fourier series can be re-written as:
$$|x| = \frac{\pi}{2}-\frac{4}{\pi}\sum_{n=0}^{\infty}\frac{\cos((2n+1)x)}{(2n+1)^2}$$

Plugging $x=0$ we get that $\displaystyle \sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$.


$\mathbf{\beta)}$ Plugging $x=0$ and using the series we derived in this topic we get that:
 $$0=\frac{\pi}{2}+\frac{2}{\pi}\left [ \sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}-\sum_{n=1}^{\infty}\frac{1}{n^2} \right ] \Leftrightarrow \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$